Variation of the longest increasing subsequence algorithm

This problem requires a solution similar to the longest increasing subsequence one. Basically on each step you either reset the current streak or increments it, adding the value to the return value. Code is down below, cheers, ACC.

Count Alternating Subarrays - LeetCode

3101. Count Alternating Subarrays

Medium

You are given a binary array nums.

We call a subarray alternating if no two adjacent elements in the subarray have the same value.

Return the number of alternating subarrays in nums.

 

Example 1:

Input: nums = [0,1,1,1]

Output: 5

Explanation:

The following subarrays are alternating: [0], [1], [1], [1], and [0,1].

Example 2:

Input: nums = [1,0,1,0]

Output: 10

Explanation:

Every subarray of the array is alternating. There are 10 possible subarrays that we can choose.

 

Constraints:

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.

Accepted

26,802

Submissions

47,966

public long CountAlternatingSubarrays(int[] nums)
{
    long retVal = 0;

    int currentStreak = 0;
    for (int i = 0; i < nums.Length; i++)
    {
        currentStreak = (i == 0 || nums[i] == nums[i - 1]) ? 1 : currentStreak + 1;
        retVal += currentStreak;
    }

    return retVal;
}